k2c901_1 wrote: A quadrilateral $PQRS$ has an inscribed circle, the points of tangencies with sides $PQ$, $QR$, $RS$, $SP$ being $A,B,C,D$, respectively. Let the midpoints of $AB$, $BC$, $CD$, $DA$ be $E,F,G,H$, respectively. Prove that the angle between segments $PR$ and $QS$ is equal to the angle between segments $EG$ and $FH$. We will work with directed angles modulo 180°. Let I be the center of the incircle of the quadrilateral PQRS. Since the incircle touches the sides SP and PQ of the quadrilateral PQRS at the points D and A, these points D and A are symmetric to each other with respect to the line PI. Thus, $DA\perp PI$, and the midpoint H of the segment DA lies on the line PI. Similarly, $AB\perp QI$, $BC\perp RI$, $CD\perp SI$, and the midpoints E, F, G of the segments AB, BC, CD lie on the lines QI, RI, SI, respectively. Since $DA\perp PI$, we have < IHD = 90°; equivalently, < IHD = - 90° (since 90° = - 90° as we are using directed angles modulo 180°). On the other hand, since the incircle of the quadrilateral PQRS, centered at I, touches the side SP at the point D, we have $ID\perp SP$ and thus < IDP = 90°. Hence, < IHD = - < IDP. Also, trivially, < HID = - < DIP. Thus, the triangles IHD and IDP are inversely similar. Therefore, $\frac{IH}{ID}=\frac{ID}{IP}$, or, equivalently, $IH\cdot IP=ID^2$. Similarly, $IF\cdot IR=IB^2$. But ID = IB, since the points D and B both lie on the incircle of the quadrilateral PQRS, which is centered at I; thus, we get $IH\cdot IP=IF\cdot IR$. Hence, by the converse of the intersecting chords theorem, it follows that the points H, P, F and R lie on one circle. Thus, < PRF = < PHF; equivalently, < (PR; RI) = < (PI; FH). Similarly, < (QS; SI) = < (QI; EG). On the other hand, since $BC\perp RI$ and $CD\perp SI$, we have < (RI; BC) = 90° and < (SI; CD) = 90°, so that < (RI; SI) = < (RI; BC) + < (BC; CD) - < (SI; CD) = 90° + < (BC; CD) - 90° = < (BC; CD) = < BCD. But similarly, < (QI; PI) = < BAD, and since the points A, B, C, D lie on one circle (namely, on the incircle of the quadrilateral PQRS), we have < BCD = < BAD. Thus, < (RI; SI) = < (QI; PI). Hence, < (PR; QS) = < (PR; RI) + < (RI; SI) - < (QS; SI) = < (PI; FH) + < (QI; PI) - < (QI; EG) = < (PI; FH) + < (EG; PI) = < (EG; FH). And the problem is solved. darij